chongo@nsc.uucp (Landon Noll) (10/15/83)
to prove that you have a 2/3 chance to win by reselecting to a friend i setup a 3 door experiment (using the Knuth 3 Vol. set as doors) where i used a die to place a large ball behind one of the three vols. "is it behind Sorting and Searching, or Semi-numerical Algor. ... ?" repeated trials showed that the answer was near the value 2/3. my friend then wrote a simulation prog. which also showed the answer was near 2/3. so from this we sat down and came up with the generalised case: the chance of winning by reselecting in an N door game is: N - 1 ------- 2 N -2*N that is: you select a door, Monty shows you a bad prize door and offers to allow you to select a new door if you like. you will win by this reselection (n-1)/(n^2-2*n) of the time. the value for n=3 yields: 2/3 here is how this value comes up: you will ONLY win by reselecting IF your first choice is BAD. since your first choice will be good 1/n of the time, your first choice will be bad 1-(1/n) = (n-1)/n of the time. when you reselect you will not select the first choice door and you will not select the door which Monty showed you. so out of n doors, you will have n-2 to reselect from. the chance of selecting a single door out of n-2 is 1/(n-2). now the prize being one of the n-2 doors only occurs (n-1)/n of the time. i.e., when you didnt select the good prize in the first place. since the 1/(n-2) chance depends of the first (n-1)/n chance we multiply them and get: (n-1)/(n^2-2*n). Note that for finite N: (N-1)/(N^2-2*N) > 1/N 1/N is the chance you will win by keeping your first choice. THUS: FOR A FINITE NUMBER OF DOORS, IT IS ALWAYS BEST TO RESELECT! chongo /\../\